Question

Out of (2n + 1) tickets consecutively numbered starting with 1, three are drawn at random. Find the chance that the numbers on them are in A.P.

Answer 1

No. of ways 3 tickets can be drawn from (2n+1) tickets = (2n+1)C(3)

Say, E = event that 3 cards drawn will have nos. in AP

There will be (2n-1) groups of 3 numbers each which are in AP

When common difference =1, groups are {(1,2,3),(2,3,4),...,(2n-1,2n,2n+1)}

When c.d. =2, groups are {(1,3,5),(2,4,6),...,(2n-3, 2n-1, 2n+1)}

When c.d. =n, groups are {(1,n+1,2n+1)}

So, total no. of group of 3 nos. drawn from (2n+1) which are in AP

= (2n-1) + (2n-3) + (2n-5) +...+ 3 + 1

= n^2

So, probability of the same = n^2 / (2n+1)C(3) = 3n/(4n^2 - 1)